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Finding the ring of integers and its algorithms in algebraic
number theory
1
Dr.K.Vijaya Kumar ,
2Dr.C.Ashok Kumar
1Associate Prof., Dept. of Mathematics
Samskruthi college of Engineering and Technology, Hyderabad (India)
2Acadamic Consultant, HOD Dept.of Mathematics
University PG Centre Kollapur, Palamuru University, Telangana (India)
ABSTRACT
We examine the algorithmic problem of finding the ring of integers of a given algebraic number field. In
practice it is often assumed that this problem is solved, but the theoretical results show that it is not manageable
for the numerical fields defined by equations with very high coefficients.Such fields occur in the number field
sieve algorithm for factoring integers. Applying a variant of a standard algorithm for finding rings of integers,
one finds a subring of the number field that one may view as the "best guess" one has for the ring of integers.
This best guess is probably often correct. Our main concern is what can be proved about this subring. We show
that it has a particularly transparent local structure, which is reminiscent of the structure of tamely ramified
extensions of local fields.
I. INTRODUCTION
In this work concerned with the following problem from algorithmic algebraic number theory: given an algebraic number field K, determine its ring of integers. The apparent contradiction is easy to resolve. Namely, all computational experience so far is limited to "small" number fields K, such as number fields that are given as
K= Q[X]/fQ[X], where Q is the field of rational numbers and f is an irreducible polynomial of small degree with small integer coefficients. The algorithms that are used for small fields will not always work when they are applied to "large" number fields. Large number fields are already making their appearance in applications of algebraic number theory, and the determination of their rings of integers is generally avoided. One can attempt to determine the ring of integer’sof a given number field K in a similarly naive manner. One starts from an order in K, i. e., a subring A of for which the index (: A) of additive groups is finite; for example, one may take A = Z [], where K is an algebraic integer with K = Q () . As we shall see, one can determine if the largest squarefree divisor m of the discriminant ΔA of A is known.
Material and Methods
Given an algebraic number field K, determine its ring of integers
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+ c0 . This ring is often denoted by OK. Since any rational integer number belongs to K and is an integral
element of K, the ring Z is always a subring of OK.
The ring Z is the simplest possible ring of integers. Namely, Z = OQ where Q is the field of rational numbers
and indeed, in algebraic number theory the elements of Z are often called the "rational integers" because of
this.The ring of integers of an algebraic number field is the unique maximal order in the field.The ring of
integers OK is a finitely-generated Z-module. Indeed, it is a free Z-module, and thus has an integral basis, that
is a basisb1, … ,bn∈ OK of the Q-vector space K such that each element x in OK can be uniquely represented
as
x = ibi
with ai∈ZThe rank n of OK as a free Z-module is equal to the degree of K over Q.
Definition 1.1. An algebraic number field is a finite algebraic extension of Q.
Definition 1.2. Let A be an integral domain, K be a field that contains A and L be an extension of K. x ∈ L is an integral element if and only if there exist
n−1, . . . ,0 ∈A such that xn + an−1xn−1 . . . + 0 = 0
In an algebraic number field, integral elements are called algebraic integers.
We will see that the integral elements form the Ring of Integers and that every element in the ring of integers can be decomposed into irreducible elements (using the Noetherian Ring property). However, uniqueness cannot always be insured. Instead, we will restrict our attention to the ideals of the Ring of Integers and demonstrate that they can be decomposed uniquely into prime ideals.
Integral Elements form a Ring. In order to show that the set of integral elements do indeed form a ring, we first need the following lemma.
Proposition 1. x∈ K is integral over A if and only if there is a finitely generated A-submodule of K such that xM⊂ M
Proof. ⇒ If xn + an−1 x n−1
+ . . . + a0 = 0 for an−1, ..., a0∈ A, then consider the A-module
N = span (1, x, ...,xn−1 )
n = 1 implies that xn ∈ N. Therefore, N is finitely generated and xN⊂ N. The others powers of x follow from
induction.
⇐ Conversely, let M = spanA(u1, . . . , un) be a finitely generated module over K. Furthermore, assume that xM⊂
M for some x∈ K. Then
xui = ai1u1 + . . . + ainun, ∀i∈ (1, . . . , n)
This leads to the following linear equations:
(x− a11)u1−a12u2 − . . . − a1nun = 0
Let B be the matrix formed by the coefficients of these linear equations. Since B has a non-zero kernel, we conclude that det(B) = 0. This implies that x satisfies an equation of the form
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where thei∈ A. This polynomial is monic since the permutation definition of the determinant shows that degree
n terms only occur when multiplying ∏bii. Thus n=1 and x is an integeral element.
Proposition: If A is an integral domain, K a field that contains A and L an extension of K, then the set
{x∈ L : ∃an−1, . . . , a0∈ A : xn + an−1xn−1 + . . . + a0 = 0}
forms a ring.
Proof. Let x and y be integral elements and M and N be finitely generated submodules of K such that xM ⊂ M
and yN⊂ N. We verify that x · y and x + y are also integral. We do so by considering the (finitely generated) submoduleMN formed by the span of all products of elements in M and N. MN is closed under multiplication by
x · y and x + y since x m ∈ M and yn∈ Nfor m∈ M and for n ∈ N.
Ring of Integers are Finitely Generated
Let K be a number field and α ∈ K, then there exists an integer multiple of α that is an algebraic integer.
Proof. By assumption, α satisfies an equation of the form
iwherebi∈ Q and bn = 1
Let i be the l.c.m of the bi. Then, multiplying by lm we have that
lmm + bn-1l(l n-1
)n-1+ ….+ b0l m
=0
Thus l is an integral element.
II. RESULT AND DISCUSSION
Algebraic and integer algebraic numbers have a useful structure. For example, they are closed by addition and multiplication:
Let and be algebraic numbers. Then (if0), , and are also algebraic numbers. Ifand are
algebraic integers, then so are and .
Let be a complex number. Then define tobe the set of all polynomials in with rational
coefficientsthat is, expressions of the forma0+ a1ak
for the some k, where all the aiarerational.
Define to be the set of all polynomials in with integers coefficients.
If is an algebraic number, then equals the set of all expressions a0 + a1ad-1dwhere
d=deg() and the ai are rational.
If is an algebraic integer, thenequals the set of all expressions a0 + a1+ .... + ad-1d-1,where the ai are integers.
PROOF
The point is that higher powers of can be "reduced" using the minimal polynomial; since d+ c
d-1d-1+ ….+ c0
=0 for some coefficients ci, d can be rewritten as a linear combination of lower powers of This can similarly
be done for d+1, d+2, and so on.
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= can be absorbed into the bterm, and so on.
The proof of the theorem uses some linear algebra to explicitly find polynomials thatand satisfy.
The idea is to look at() or Z[], and establish a dependency relation between the
powers of (or the powers of ). Rather than developing the machinery necessary to do
the proof in general, it is easiest to look at an explicit example.
Find the minimal polynomial of + .
Consider the powers of + :
Setting some linear combination of these powers equal to 0 yields four linear equations in five unknowns, this always have a nontrivial solution. In this case, it’s not hard to see that So the minimal
polynomial isx4-10x2+1.
III. CONCLUSION
We conclude that there are well-defined operations of addition and multiplication on the set Znof congruence
classes of integers modulo n:the sum of the congruence classes of integers x and y is the congruence class of
x+y, and the product of these congruence classes is the congruence class of xy. These operations of addition and multiplication on congruence classes do not depend on the choice of representatives of those congruence classes.
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